$2\sqrt[3]{x^2}\cdot \sqrt{16x}\\\\=2\sqrt[3]{x^2} \cdot 4\sqrt x\\\\=8x^{\tfrac 23} \cdot x^{\tfrac 12}\\\\=8\cdot x^{\tfrac 23 + \tfrac 12}\\\\=8\cdot x^{\tfrac 76}\\\\=8\sqrt[6]{x^7}\\\\=8\sqrt[6]{x^6 \cdot x}\\\\=8x\sqrt[6]{x}$

$\text{Hence the answer is A}$

4 0

1 year ago

Read 2 more answers

If 8,a,b,27 are in geometric sequence, find the value of a and b.

wel

Answer:

a = 12 and b = 18

Step-by-step explanation:

Given that,

8,a,b,27 are in geometric sequence.

For a GP, the nth term is given by :

$a_n=ar^{n-1}$

Put n = 4

$a_4=ar^{4-1}\\\\a_4=ar^3\\\\27=8\times r^3\\\\r^3=\dfrac{27}{8}\\\\r=\dfrac{3}{2}=1.5$

Put n = 2,

$a_2=ar^{2-1}\\\\a=ar\\\\a=8\times 1.5\\\\a=12$

Put n = 3

$a_3=ar^2\\\\=8\times 1.5^2\\\\b =18$

So, the values of a and b is 12 and 18 respectively.

3 0

2 years ago